Problem: Let $x_1< x_2 < x_3$ be the three real roots of the equation $\sqrt{2014} x^3 - 4029x^2 + 2 = 0$. Find $x_2(x_1+x_3)$.
The expression $x_2(x_1+x_3)$ is not symmetric in the roots $x_1, x_2, x_3,$ so Vieta's formulas can't be used directly to find its value. We hope that we can determine some of the values of the roots explicitly. Letting $a = \sqrt{2014},$ the equation becomes \[ax^3 - (2a^2+1)x^2 + 2 = 0.\]We can rearrange this as \[(ax^3-x^2) - (2a^2x^2-2) = 0,\]or \[x^2(ax-1) - 2(ax-1)(ax+1) = 0.\]Therefore, we have \[(ax-1)(x^2-2ax-2) = 0.\]It follows that one of the roots of the equation is $x = \tfrac{1}{a},$ and the other two roots satisfy the quadratic $x^2 - 2ax - 2 = 0.$ By Vieta's formulas, the product of the roots of the quadratic is $-2,$ which is negative, so one of the roots must be negative and the other must be positive. Furthermore, the sum of the roots is $2a,$ so the positive root must be greater than $2a.$ Since $2a > \tfrac1a,$ it follows that $\tfrac{1}{a}$ is the middle root of the equation. That is, $x_2 = \tfrac1a.$

Then $x_1$ and $x_3$ are the roots of $x^2-2ax-2=0,$ so by Vieta, $x_1+x_3=2a.$ Thus, \[x_2(x_1+x_3) = \frac1a \cdot 2a = \boxed{2}.\]